The second order derivative of which of the following functions is $5^{\mathrm{x}}$ ?

We need to find a function that, when differentiated twice, gives us $5^{x}$.

What does "second order derivative" mean? It means we take the derivative twice. So if our function is $y$, we first find $y'$ (first derivative), then find $y''$ (second derivative). We want $y'' = 5^x$.

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The derivative of $a^{x}$ is:

$\frac{d}{dx}(a^{x}) = a^{x} \cdot \ln{a}$

When you differentiate something like $5^x$, you get $5^x \cdot \ln{5}$.

Note: $\ln{5}$ is the same as $\log_{e}{5}$ (natural logarithm of $5$). It's just a number (approximately $1.609$).

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The Smart Approach: Work Backwards!

We want $y'' = 5^x$. Each time we differentiate $5^x$, we multiply by $\ln{5}$. So each time we undo a differentiation, we divide by $\ln{5}$.

Working backwards:

$y'' = 5^x$

$y' = \frac{5^x}{\ln{5}}$ (divide by $\ln{5}$ once)

$y = \frac{5^x}{(\ln{5})^2}$ (divide by $\ln{5}$ twice)

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Let's verify our answer by differentiating twice.

Starting with $y = \frac{5^{x}}{(\ln{5})^{2}}$

First derivative:

$y' = \frac{1}{(\ln{5})^{2}} \cdot \frac{d}{dx}(5^x) = \frac{1}{(\ln{5})^{2}} \cdot 5^{x} \cdot \ln{5} = \frac{5^{x}}{\ln{5}}$

Second derivative:

$y'' = \frac{1}{\ln{5}} \cdot \frac{d}{dx}(5^x) = \frac{1}{\ln{5}} \cdot 5^{x} \cdot \ln{5} = 5^{x}$

It works! The $\ln{5}$ terms cancel out perfectly.

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Why the other options don't work:

If $y = \frac{5^{x}}{\ln{5}}$, you'd get $y'' = 5^x \cdot \ln{5}$ (extra $\ln{5}$ left over)

If $y = 5^{x} \cdot (\ln{5})^{2}$, you'd get $y'' = 5^{x} \cdot (\ln{5})^{4}$ (way too many $\ln{5}$ terms)

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Hence, the answer is:

$\boxed{y = \frac{5^{x}}{(\ln{5})^{2}}}$